Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int n, m, Q;
double sy, l[200100], r[200100], sum[200100], zl, zr, x, y, ans;
struct Node {
  double x, y;
} bj[2][200100], tmp, L, R;
inline bool lef(Node u, Node v, Node w) {
  if (u.x == v.x) return w.x < u.x;
  double t;
  t = v.x - (v.x - u.x) * (v.y - w.y) / (v.y - u.y);
  return w.x < t;
}
inline double ask(int u) {
  double p, q;
  if (bj[0][u].x == x)
    p = x;
  else {
    p = x - (x - bj[0][u].x) * (y - sy) / y;
  }
  if (bj[1][u].x == x)
    q = x;
  else {
    q = x - (x - bj[1][u].x) * (y - sy) / y;
  }
  return max((double)0, min(q, zr) - max(p, zl));
}
int main() {
  int i, j, p, q, mid, a, b;
  cin >> sy >> zl >> zr >> n;
  L.y = R.y = sy, L.x = zl, R.x = zr;
  for (i = 1; i <= n; i++) {
    scanf("%lf%lf", &l[i], &r[i]);
    sum[i] = sum[i - 1] + r[i] - l[i];
    bj[0][i].y = 0, bj[0][i].x = l[i];
    bj[1][i].y = 0, bj[1][i].x = r[i];
  }
  cin >> m;
  for (i = 1; i <= m; i++) {
    scanf("%lf%lf", &x, &y);
    ans = 0;
    tmp.x = x, tmp.y = y;
    for (p = 1, q = n; p < q;) {
      mid = ((p + q) >> 1);
      if (lef(tmp, bj[0][mid], L))
        q = mid;
      else
        p = mid + 1;
    }
    a = p - 1;
    for (p = 1, q = n; p < q;) {
      mid = ((p + q) >> 1);
      if (!lef(tmp, bj[1][mid], R))
        p = mid + 1;
      else
        q = mid;
    }
    b = p;
    printf("%.10f\n", ask(a) + ask(b) * (a != b) +
                          max((double)0, sum[b - 1] - sum[a]) * (y - sy) / y);
  }
}